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Hooke's Law Calculator

Find the force and energy in a stretched spring.

How it is calculated

Hooke's law says the force of a spring is proportional to its extension: F = k × x, where k is the spring constant in newtons per metre (N/m) and x is the stretch or compression in metres. The tool also gives the stored elastic energy PE = ½ × k × x² in joules. The force is shown as a magnitude, so the sign of x only tells direction and is dropped from the displayed force, while the energy stays positive. The spring constant must be greater than zero. The law holds only within the elastic limit of the spring; past it, the spring deforms permanently and the formula no longer applies.

Worked example

A spring with k = 200 N/m is stretched by 0.05 m (5 cm). Force F = 200 × 0.05 = 10.00 N. Energy PE = ½ × 200 × 0.05² = 100 × 0.0025 = 0.2500 J.

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Common mistakes

  • Entering the stretch in centimetres or millimetres instead of metres: 5 cm is 0.05.
  • Using the total length of the spring instead of the change from its rest length.
  • Doubling the stretch and expecting double the energy: energy grows with the square, so it rises fourfold.
  • Using a k in N/mm or kgf/cm without converting to N/m.
  • Applying the law beyond the elastic limit, where force no longer grows in proportion.

About this calculator

This calculator applies Hooke's law, F = kx, to find the restoring force a spring exerts when stretched or compressed by a distance x, where k is the spring constant (stiffness); it also computes the elastic potential energy stored, PE = ½kx². Hooke's law is linear and only holds within a spring's elastic limit — beyond that, the material deforms permanently and the relationship breaks down.

Physics and mechanical engineering students use it for spring and simple-harmonic-motion problems, while mechanical and product designers apply it to size springs in suspensions, scales, mattresses and mechanical assemblies.

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Comments (1)

  • Editorial team

    If you do not know a spring's constant, measure it: hang a known mass, take its weight as mass times 9.81 N/kg, and divide by the stretch in metres. A 0.5 kg load that stretches the spring 0.049 m gives about 100 N/m. Two identical springs side by side double the stiffness, while the same two in a chain halve it. Which spring are you analysing, a toy, a suspension or a lab sample?

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